Q21 - Cognitive Psychologist: The majority of

[opulence2001]

For this question I choose answer C. I'm figuring that this is true but not necessary for the conclusion to be drawn.

E is the correct answer but I'm not making the connection as to how it leads me to draw the conclusion. I know this is a formal logic question, but this one is not connecting for me.

[noah]

Wow, this is a doozy to explain! Let's get into it!

I wouldn't necessarily consider this a conditional logic question, since the premises are about mosts and somes. But, there are rules to those that apply here.

Here's what we're told:

Skill Artist --(most)--> Creat.People --> Abst. Reason
Skill Artist --(some)--> NOT Famous

Let's start with the first premise. What can we infer about people with Abstract Reasoning. When inferring up a must be true arrow, all we could say is that "Some people who are good at Abstract Reasoning are Creative People" (the reverse of an All is a Some).

Continuing up the chain, we can only say that there are "Some people who are good at Abstract Reasoning that are Skilled Artists" (the reverse of a most is a some).

So, how do we get to this: Good Abst. Reason --(some)--> Famous

We'd have to know that there are some Creative People that are famous. It'd be great if we knew that all Skilled Artists were famous because then we'd know there's a famous one among the Creative People group, however we actually know that some of the Skilled Artists are NOT famous. So, let's reset our view on confirming that there are some famous Creative People. To get to that, it's good to know this rule "most + most = some"

Take this example:

Most kids like ice cream.
Most kids like running.

We can infer that there must be at least one kid who likes both. (To prove it, imagine there were 10 kids in the world, and you'll see that it's at least 6 kids who like ice cream and 6 who like running, so there must be an overlap).

So, if we knew that Most Skilled Artists are famous, then we'd have this:

Most Skill Artist are Famous.
Most Skill Artist are Creative.

We could infer that there's at least one Famous Creative person.

If we knew that, then we would have this chain:

Famous --(some)--> Creat.People --> Good Abst. Reason

And we could reverse infer up the original chain with a "some", giving us the conclusion:

Good Abst. Reason --(some)--> Creat.People -(some)-> Famous

(C) is somewhat easily eliminated because it's a "premise-booster." We already know that from the second sentence!

Tough question. Does that help?

#officialexplanation

[opulence2001]

Thank-you so much for tackling this question. I went back to it after a few weeks and I understand why B is wrong, and I was going to ask why A is wrong. However, as I was typing my question I think I finally figured it out.

S = skilled artists, R= good at abstract reasoning, C= creative ppl, and F= famous.

I went back to your explanation and I have this:

S(most) -> C(all)--> R(all)

which in inverse work out as:

R(some) --> C(some) --> S(all)

Conclusion: R(some) --> F

I just continued along the chain of the inverse to get the assumption. The premise not all skilled artists are famous I think is just thrown in to throw ppl off.

So, I figured it would have to be assumed S(most) --> F in order for this argument to work. It would look like the following:

R(some) --> C(some) --> S(all)--> F

It follows that R-->F

The part in red is what would have to be assumed.

Is this thinking correct?

[noah]

Nice work. It looks like this part is off

S(most) -> C(all)--> R(all)

which in inverse work out as:

R(some) --> C(some) --> S(all)

The inverse of S-(most)-> C is actually C-(some)-> S . In short, most and some reversed become some. BTW, it might be easier for you to simply write the "some" or "most" above the arrow if you use this sort of notation. There's only one modifier per relationship.

I agree that the "not all skilled artists are famous" is a distraction.

The reasoning is basically that you need S -(most)-> R (because S-(most)-> C-(all)-> R) and S -(most)-> F to ensure an overlap between R and F.

I believe you've incorrectly written that S-(all)-> F, all we know is that S-(most)-> F.

The key is to know a few some and most rules:

some + some = ?
some + most = ?
most + most = overlap

With this sort of question, it's totally possible that I messed up, so fire away!

[hwsitgoing]

Hello,

I'm sorry, but I've read over this thread several times and still don't quite understand how you reached the correct answer. Is there any way you could re explain how you arrived at E in a very simple and methodical way? I can't believe the lsat would have a question this time consuming!

I understand the inverse process and formal logic concepts but now how you make the jump from the inverse to the correct answer.

Thank you!

[noah]

I guess the simplest explanation would be this:

We learn that SA --(most)--> CP + AR

And we need to conclude that SA --(some)--> F

To get from X --(most)--> Y to Y --(some)--> Z, we can expect the LSAT to provide X --(most)--> Z. With two most statements about the same entity, X, we can infer an overlap in the two characteristics (Z and Y).

Back to the artists! We want another most statement about SA that will give us an overlap between F and AR, and it should be, as (E) is, SA --(most)--> Famous

So, in conclusion, the slick approach is to see we get a most statement about the characteristics that most Skill Artists have, and end with a some statement about one of those characteristics (Good at abstract reasoning) and another characteristic (Famous), so we need one more most statement about Skill Artists being Famous.

Does that clear it up?

[Shiggins]

Noah, I liked the rules that you placed with the categorical logic. I just have one question. I see how most to most overlaps but is that only when the arrows are in different directions
example above:

F <- SA-> AR

SA= skilled artist
AR= Abstract reasoning
F= Famous

I believe most to most when going in the same direction can not be done.
Ex.
Most tall people have knee problems. Most people with knee problems have knee surgery. I do not think you can infer Most Tall people have knee surgery.

Your explanation on this question was very good, I just want to make sure what I typed is correct, otherwise if you can explain where I went wrong. Thank you.

[noah]

I believe most to most when going in the same direction can not be done.
Ex.
Most tall people have knee problems. Most people with knee problems have knee surgery. I do not think you can infer Most Tall people have knee surgery.

You got it! Another way to think of it is that both most statements have to start with the same thing. Most X have Y and Most X have Z.

[syousif3]

okay so I thought answer E would better if it has some instead of most.

Most SA--> C (all)-->AR(some)-->F

so we can infer that

some AR + some C --->SA

i'm just not sure how you can infer most SA are famous, unless you can also infer that all most SA are C.. and Some C are F. I'm so confused can someone please help?

[timmydoeslsat]

I would personally recommend that you alter the way you setup your argument to look for the sufficient assumption.

Instead of using arrows with quantifying statements, I would recommend using words. Save the arrows for true conditional statements.

Would you agree that our argument is structured like this?

SA most VCP ---> GAR
SA some ~F
__________________
GAR some F


We want to prove this conclusion true, no matter how strong the additional premise is that we supply to this argument.

We do see that we have a positive case of F being used, something that is absent from evidence. If we supply the argument with another some statement, there is no way we are going to be able to infer anything. We cannot combine most and some statements, or two statements together. However, we can infer something when we have two most statements concerning the same variable.

A most B
A most C

There must be an overlap of B and C. We can infer that B some C from the two statements above.

In this argument, we have:

SA most VCP ---> GAR

So if we plug in that SA most F, you will have the overlap of VCP and F. Since we would have F some VCP, we can infer that F some GAR. This is due to us knowing from the premises that ever VCP gives us GAR.

[patrice.antoine]

Ok I will try:

My conditionals are:

Key Symbols
SA - Skilled Artist
VC - Very Creative
GAR - Good Abstract Reasoning
F - Famous

Breaking down each sentence:

SA --(Most)-->VC
VC --> GAR
SA --(Some)--> ~F
___________________
Conclusion: GAR--(Some)--> F


We can combine the first two conditionals:

Premise 1: SA --(Most)-->VC--> GAR
Premise 2: SA --(Some)--> ~F
___________________
Conclusion: GAR--(Some)--> F


So if we know that MOST skilled artists are very creative and as such, good at abstract reasoning and that SOME skilled artists are not famous how can we then conclude that SOME of those good at abstraction are famous?

By adding a "MOST" statement connecting the gap between famous and skilled artists:

Premise 1: SA --(Most)-->VC--> GAR
Premise 2: SA --(Some)--> ~F
Premise 3: SA--(Most)-->F
___________________
Conclusion: GAR--(Some)--> F

So if we know that MOST SAs are GAR and that MOST SAs are famous can we then conclude that at least one (SOME) GARs are famous?

Yes, we certainly can.

You can also try plugging in the other 4 answer choices as premises to see if they can help draw us to our conclusion. They do not.

HTH!!

[noah]

okay so I thought answer E would better if it has some instead of most.

Most SA--> C (all)-->AR(some)-->F

so we can infer that

some AR + some C --->SA

i'm just not sure how you can infer most SA are famous, unless you can also infer that all most SA are C.. and Some C are F. I'm so confused can someone please help?

I think you may be confused about your job here. You're trying to make the argument presented work by choosing an assumption. You're not inferring the answer. Most is stronger than some, so in terms of something that makes the argument valid, it's a stronger answer in this case.

[samuelfbaron]

I understand the logic chain but I don't understand how choice (E) closes the gap in the reasoning?

Most S.A. --> CR
All CR --> AR

Some AR --> Famous

SA --> CR --> AR --> Famous

I'm not sure how (E) closes the gap.

[noah]

I understand the logic chain but I don't understand how choice (E) closes the gap in the reasoning?

Most S.A. --> CR
All CR --> AR

Some AR --> Famous

SA --> CR --> AR --> Famous

I'm not sure how (E) closes the gap.

I don't see the logic chain like that. I see it like this, using your terms (pieces numbered for ease of discussion):

1. SA --most--> CR
2. CR --> AR

3. Which allows us to say that SA --most--> AR

4. However, SA --some--> NOT Famous

5. Conclusion: AR --some--> Famous

6. If we know (E), that SA --most--> Famous, then we have a most+most overlap: most SA are AR and most SA are Famous.

7. This means that there must be at least one SA that is both AR and Famous, which tells us that there are some people (at least one) that are good at abstract reasoning (AR) that are famous, which is the conclusion.

Make sense?

[Ibrahim.diallo]

I got this question wrong under time condition but when reviewing it, I thought it could be really easy to solve actually. I want to make sure my thought process is solid.

SA=Skilled Artists
CP=Creative People
AR=Abstract Reasoning
F=Famous

Premises:
SA-----(most)--->CP------->AR

SA----(some)--->~F

Conclusion:
AR-----(some)---->~F

I reasoned that no matter what, the answer choice has to say something about Famous AND Skilled Artist (or Not Famous & Not Skilled Artists). With this logic, you can eliminate A, B & D.

Now down to E and C. C simply repeats one of the premises of the argument - "some skilled artists are not famous" is the same thing as "not all skilled artists are famous". So I got rid of it for that reason. Then was left with E - which is the correct answer for reasons stated previously on this thread.

Did I get lucky or is this sound?

[maryadkins]

Totally sound!